bjt time analysis

Adib AbrishamifarEE Department

IUST

Lecture 6 – BJT Inverter Switching Times

Digital Integrated Circuit Design

( )oV t

( )iV tot

5 sµ5

t1

2t

4t

3t5t

6t

o

5.0 1

o

IUST: Digital IC Design LLECTURE 6 ECTURE 6 : : BJT Inverter Switching Times Adib Abrishamifar 20082/30

} BJT Inverter} Cutoff Region (to → t1)} Active Region (t1 → t2)} Saturation Region (t3 → t4) } Active Region (t4 → t5)} Recovery Region (t5 → t6)} Propagation Delay Time

} Definition of a parameter} Power Delay Product} Summary

Contents


BJT Inverter

5+

CR K=1( )oV t

BR K=10

( )iV t

BE(on)

BE(sat)

CE(sat)

jeo

e

e

V = 0.7V

V = 0.8V

V = 0.1V

C = 0.3PF

= 0.9Vm = 0.5Φ

F

BF

jc

c

c

s

o

= 20nse

= 0.2nsec= 14nsec

C = 0.15PF

= 0.7Vm =

c

0.33

ττ

τ

Φ

Delay TimeFall TimeSaturation TimeRise TimeFinal Recovery Time

o d

f

s

r

fr

t t tt t tt t tt t tt t t

− = =− = =− = =

− = =− = =

1

2 1

4 3

5 4

6 5

( )oV t

( )iV tot

5 sµ5

t1

2t

4t

3t5t

6t

o

5.0 1

o




BJT Inverter Switching Time


F R B VE VC

BE o

BE 1

BC o

BC 1

dQ ,Q i (t) = (Q + Q )dt

V (t ) = 0VV (t ) = 0.7VV (t ) = 0 -5 = -5VV (t ) = 0.7 -5 = -4.3V

o≈ ⇒

Cutoff Region (to → t1)

Ei

Bi VCQ

VEQ

Ci



( )

[ ] [ ]

B o

B 1

B

B

B

5 - 0i (t ) = 0.5mA10K5 - 0.7i (t ) = 0.43mA10K

0.5 + 0.43i = 0.465mA2

di (t) = Q Qdt

i (t)dt = Q ( ) Q ( ) Q ( ) Q ( )o

VE VC

t

VE VE o VC VC ot

t t t t

=

=

=

+ ⇒

− + −∫1

1 1

+5

CR K=1

( )oV tBR K=10

( )iV t



( )( )

1/2 1/2

VE

Q Q ( ) Q ( ) .

( )( )( )

-0.3×0.9 0.7 01- 1- 0.408PF0.7 - 0 1 2 0.9 0.9

ΔQ = 0.408PF×0.7V = 0.285PC

e e

VE VE VE o eq BE

m m

e BE oBEeq jeo

BE e e e

t t C V

V tV tC CV m

− −

∆ = − = ∆

Φ = − − − − ∆ − Φ Φ = − ≈

1

1 111 1

1

+5

CR K=1

( )oV tBR K=10

( )iV t



( )( )

( )1

o

2/3 2/3

VCt

t

1 o d

Q .

0.15×0.7 4.3 51+ 1+ 0.076PF-4.3 + 5 2 3 0.7 0.7

ΔQ = 0.076PF×0.7V = 0.053PC

0.456mA dt = 0.285 + 0.053 = 0.338PC

0.338PCt - t = t = = 0.73nsec0.465mA

VC eq BC

eq

C V

C

′∆ = ∆

′ = − − ≈

⇒

∫

( )oV t

( )iV t oot

sµ55

t1

t2

t4

t3t5 t6

o

5.0 1


Active Region (t1 → t2)

FB F VC VE

BF

BE(on) BE(sat)

CE CC CE CEsat

BE 2

BC 2

Q di (t) = + (Q + Q + Q )τ dt

V V

V = V V = VV (t ) = 0.8VV (t ) = 0.8 - 0.1 = 0.7V

→

→Ci

Ei

Bi

BEVFQ

FQtfτVCQ

VEQ



( )( )

2 2

1

B 1

B 2

B

FB F VC VE

BFt

B FBF t

2

2

1

5 - 0.7i (t ) 0.43mA10K5- 0.8i (t ) 0.42mA10K

0.43+ 0.42i 0.425mA2

Q di (t) = + (Q + Q + Q )τ dt

i (t)dt Q (t)dt Q Q Qτ

Q Q ( ) Q ( )Q ( ) 4.9 0.2 sec 0.98Q ( ) 0 ,

t

F VE VCt

F F F

F C F

F C

t tt I τ mA n PCt I

= =

= =

= =

= + ∆ + ∆ + ∆

∆ = −

= = =

=

∫ ∫1

1

1

( ) 0 Q 0.98Ft PC= ⇒ ∆ =1

+5

CR K=1

( )oV tBR K=10

( )iV t


( )( )

( )( )

1/2 1/2

VE

2/3 2/3

Q Q ( ) Q ( ) .

-0.3 0.9 0.8 0.71- 1- 0.745PF0.8 - 0.7 1 2 0.9 0.9

ΔQ = 0.745PF 0.1V = 0.075PCQ .

0.15 0.7 0.7 4.31 1+0.7 4.3 2 3 0.7 0.7

VE VE VE o eq BE

VC eq BC

eq

t t C V

C V

C

∆ = − = ∆

× = − =

×′∆ = ∆

× ′ = − − − +

1

( )2

1

VC

t

t

2 1

By neglecting the recombination ter

0.117PF

ΔQ = 0.1176PF 5.0V = 0.584PC

0.425mA dt 0.98 0.075 0.584 1.64PC

1.64 3.9 sec0.42

m

5

:

fPCt t t nmA

=

×

= + + =

⇒ − = = =

∫


( )oV t

( )iV t oot

sµ55

t1

t2

t4

t3t5 t6

o

5.0 1



FQ

.0 98

t10

t2 t

} Now to include the recombination term, we assume a linear increase in QF from t1 to t2

} Note: The loss of carrier by recombination results in an increase in the fall time

( ) ( )( )

( )

2

1

t2 1

t

2 1

0.98PC0.425mA dt 1.64PC

sec 2

1.64 4.2 sec0.425- 0.035f

t tn

PCt t t nmA

−= × +

⇒ − = = =

∫1

14


} BJT Inverter} Cutoff Region (to → t1)} Active Region (t1 → t2)} Saturation Region (t3 → t4)} Active Region (t4 → t5)} Recovery Region (t5 → t6)} Propagation Delay Time




} Input pulse is wide and td + tf = 4.9ns then transistor reaches to steady state saturation

5

t3o

BFi

BRi . V0 8

Saturation Region (t3 → t4)

3

3

( )

:

( ) ( )

( ) ( ) ( ) ( )

( )

s

C s

F s

C s s

F s

tCs s BR

F

C Cs s BF s BR

F F

s BF B

BF

R

BR

I Q I

Step in tI Q dQ

dtIQ t i Ke

I II Q t i i K

K

i

i

i i

τ

β τ

β τ

τβ

τ τβ β

τ

−

− =

− = +

= − +

⇒ = − = − +

= −


Saturation Region (t3 → t4)

( )

( )

( )

4 3

( ) ( )

5 - 0.8 0.4210K0 - 0.8 0.0810K

14 700.20.42 0.0820 24 sec

4.9 70 0.08

stCs s BR BF BR

F

s

BF BRs s

C F BR

BF

BR

BFF

F

s

IQ t i i i e

t t Q oi it t t Ln

I i

i mA

i mA

t Ln n

ττβ

τβ

τβ

τ

− = − + −

= ⇒ → ⇒

−− = =

−

= =

= = −

= = =

+= =

+

4

( )oV t

( )iV t oot

sµ55

t1

t2

t4

t3t5 t6

o

5.0 1



5 5

4

FB F VC VE

BFt

B FBF t

B 4

B 5

B

Q di (t) = + (Q + Q + Q )τ dt

i (t)dt Q (t)dt Q Q Qτ

During Turn Off Q Q (During Turn-on)0 - 0.8i (t ) 0.08mA10K

0 - 0.7i (t ) 0.07mA10K

i 0.075mA

t

F VE VCt

tot tot

= + ∆ + ∆ + ∆

∆ = −∆

= = −

= = −

= −

∫ ∫4

1



5

5 4 r

R

5 4 r

By neglecting recombination term :

(-0.075mA)dt 1.64

1.64 22 sec0.075

Including the recombination termand assuming a linear decrease of Q :

1.64 15 sec(0.075 + 0.035)

No

t

t

PC

PCt t t nmA

PCt t t nmA

= −

−⇒ − = = =

−

−− = = =

−

∫4

te that the recombination can be asubstantial aid in the removal of charge from the base.

( )oV t

( )iV t oot

sµ55

t1

t2

t4

t3t5 t6

o

5.0 1


} This situation is similar to the initial delay time prior to turn on

B 5

B 6

B

6 5 f r

0 - 0.7i (t ) 0.07mA10K

i (t ) 0mA

i 0.035mAQ Q 0.338PC

-0.338PCt - t t 9.7 sec-0.035mA

VC VE

n

= = −

=

= −∆ + ∆ = −

= = =

Recovery Region (t5 → t6)

( )oV t

( )iV t oot

sµ55

t1

t2

t4

t3t5 t6

o

5.0 1


Propagation Delay Time

tVout

Vin

t

50%

tpHL

50%

tpLH

tf

90%

10%

tr

PHL d f

PLH s r PHL

PHL PLHP

t = t t /2 0.73 4.2/2 2.8nsect t t /2 24 15/2 31.5nsec >> t

t t 2.8 31.5t 17.15 sec2 2

n

+ = + == + = + =

+ += = =

} This indicates that the saturation time is a very large fraction of the total time


Definition of a parameter

?(Before) Q , Q

, Excess base current(Now) Q , Q

Q (In saturation)

Q Q Q Q

Q Q

By manipulating t

s

F CF tF R ER tR

B Bsat BS BS

a Csat tF s BS s

a

F R R RC

tF tR BR

s sBS

s

I II I I I

I Iddt

didt

dIdt

ττ τ

τ τ

τ τ τ

τ

=

= =′ ′= + =

′= =

=

= − − −

′ = +

0

hese equations:( )

0.98 , 0.5 , 0.2 sec , 20 sec 20 sec

F tF R tRs

F R

F R tF tR sn n n

α τ α ττ

α αα α τ τ τ

+⇒ =

−= = = = ⇒ ≈

1

0 w

CBE

FQ

RQ

ForwardReverse

0 w

CBE

activeQ

SQ

ForwardReverse


Power Delay Product

( ) ( )Pd PHL PLH1 1t = t t 2.8 31.5 17.15 sec2 2

0

Calculated for connecting the output to only one input

5 0.8 10 0.8 4.6181 10

dL d Vin o

dH cc Csat IH IH

IH

KIH K K

n

P P

P V I V IV

V V

=

+ = + =

= =

= +=

−⇒ = × + =

+


( )

5 0.2 4.8

0.8 0.381810

5 4.8 4.618 0.381825.76

12.882

Csat K

IHIH K

V mA V mAdisH

disH

d dL dH

d

I mA

VI mA

PP mWP P P

P mW

−= =

−= =

= × + ×

⇒ =

⇒ = +

=

1

1 2

Power Delay Product

} This is very high.


} In this lecture the switching times of the bipolar transistor were first described

} Delays in all of mode were calculated} Saturation time is very large fraction of total

times} Propagation delay time was calculated} Finally power delay product was calculated which

was very high

Summary

bjt time analysis

Documents

digital ic design

propagation delay time

bjt invertercutoff region

dt dq

phl

recombination term

2n sec

recovery region